Q: Q4. A plane electromagnetic wave travels in vacuum along z-direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is

This question has two parts:\n\n**1. Directions of Electric and Magnetic Field Vectors**\n\nFor a plane electromagnetic wave, the electric field vector (E), the magnetic field vector (B), and the direction of wave propagation are mutually perpendicular to each other. This is a key characteristic of the transverse nature of electromagnetic waves.\n\nGiven that the wave travels along the z-direction:\n* The electric field vector (E) and the magnetic field vector (B) must lie in the plane perpendicular to the z-direction, which is the **x-y plane**.\n* The vectors E and B are also perpendicular to each other.\n* The direction of propagation is given by the direction of the cross product E × B. So, if E is along the x-axis, B must be along the y-axis for the wave to propagate along the z-axis (since î × ĵ = k̂).\n\n**2. Calculation of Wavelength**\n\nThe relationship between the speed of light (c), frequency (f or ν), and wavelength (λ) is given by:\nc = fλ\n\nGiven:\nFrequency, f = 30 MHz = 30 × 10⁶ Hz\nSpeed of light in vacuum, c = 3 × 10⁸ m/s\n\nWe can find the wavelength (λ) by rearranging the formula:\nλ = c / f\nλ = (3 × 10⁸ m/s) / (30 × 10⁶ Hz)\nλ = (3 × 10⁸) / (3 × 10⁷) m\nλ = 10 m\n\nTherefore, the wavelength of the electromagnetic wave is 10 meters.

Q: Q6. A plane electromagnetic wave travels in vacuum along z-direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is

### Directions of Electric and Magnetic Field Vectors\nFor a plane electromagnetic wave traveling in vacuum along the z-direction, the electric field vector (E) and the magnetic field vector (B) must be perpendicular to each other and also perpendicular to the direction of wave propagation (z-direction). This means that E and B lie in the x-y plane. \n\nSpecifically, if the electric field oscillates along the x-axis, then the magnetic field will oscillate along the y-axis, and vice-versa. The direction of propagation of the electromagnetic wave is given by the cross product E × B.\n\n### Wavelength Calculation\nGiven:\nFrequency, ν = 30 MHz = 30 × 10⁶ Hz\nSpeed of light in vacuum, c = 3 × 10⁸ m/s\n\nThe relationship between speed of light (c), frequency (ν), and wavelength (λ) is given by:\nc = νλ\n\nTherefore, the wavelength λ can be calculated as:\nλ = c / ν\nλ = (3 × 10⁸ m/s) / (30 × 10⁶ Hz)\nλ = (3 × 10⁸) / (3 × 10⁷) m\nλ = 10 m\n\nThe wavelength of the electromagnetic wave is 10 meters.

Q: Q8.10. Suppose that the electric field amplitude of an electromagnetic wave is E₀ = 120 N/C and that its frequency is ν = 50.0 MHz. (a) Determine, B₀, ω, k, and λ. (b) Find expressions fo

Given:\nElectric field amplitude, E₀ = 120 N/C\nFrequency, ν = 50.0 MHz = 50.0 × 10⁶ Hz\nSpeed of light, c = 3 × 10⁸ m/s\n\n**(a) Determination of B₀, ω, k, and λ**\n\n1. **Magnetic Field Amplitude (B₀):**\n The relation is B₀ = E₀ / c.\n B₀ = (120 N/C) / (3 × 10⁸ m/s) = 40 × 10⁻⁸ T = 4.0 × 10⁻⁷ T (or 400 nT).\n\n2. **Angular Frequency (ω):**\n The relation is ω = 2πν.\n ω = 2π × (50.0 × 10⁶ Hz) = 100π × 10⁶ rad/s = π × 10⁸ rad/s.\n ω ≈ 3.14 × 10⁸ rad/s.\n\n3. **Wavelength (λ):**\n The relation is λ = c / ν.\n λ = (3 × 10⁸ m/s) / (50.0 × 10⁶ Hz) = (300 / 50) m = 6.0 m.\n\n4. **Wave Number / Propagation Constant (k):**\n The relation is k = 2π / λ.\n k = 2π / 6.0 m = π/3 rad/m.\n k ≈ 1.05 rad/m.\n (Alternatively, k = ω/c = (π × 10⁸ rad/s) / (3 × 10⁸ m/s) = π/3 rad/m).\n\n**(b) Expressions for E and B**\n\nLet's assume the wave propagates along the positive x-axis. The electric field E can be taken along the y-axis, and the magnetic field B along the z-axis, as they must be mutually perpendicular.\n\nThe general expressions for a plane electromagnetic wave are:\nE(x, t) = E₀ sin(kx - ωt) ŷ\nB(x, t) = B₀ sin(kx - ωt) ẑ\n\nSubstituting the calculated values:\n\n**Expression for Electric Field (E):**\nE(x, t) = 120 sin( (π/3)x - (π × 10⁸)t ) ŷ (N/C)\nOr using approximate values:\nE(x, t) = 120 sin(1.05x - 3.14 × 10⁸t) ŷ (N/C)\n\n**Expression for Magnetic Field (B):**\nB(x, t) = (4.0 × 10⁻⁷) sin( (π/3)x - (π × 10⁸)t ) ẑ (T)\nOr using approximate values:\nB(x, t) = (4.0 × 10⁻⁷) sin(1.05x - 3.14 × 10⁸t) ẑ (T)

Question 1

Q1. Figure 8.6 shows a capacitor made of two circular plates each of radius 12 cm, and separated by 5.0 mm. The capacitor is being charged by an external source (not shown in the figur

Accepted Answer

Here are the solutions to the three parts of the question:

**(a) Capacitance and Rate of Change of Potential Difference**

**1. Capacitance (C):**
The capacitance of a parallel plate capacitor is given by the formula C = (ε₀A)/d, where A is the area of the plates and d is the separation between them.
Given:
Radius, r = 12 cm = 0.12 m
Separation, d = 5.0 mm = 5.0 × 10⁻³ m
Area of circular plates, A = πr² = π(0.12)² m² ≈ 0.0452 m²
Permittivity of free space, ε₀ = 8.854 × 10⁻¹² F/m

C = (8.854 × 10⁻¹² × 0.0452) / (5.0 × 10⁻³) 
C ≈ 80.1 × 10⁻¹² F = 80.1 pF

**2. Rate of Change of Potential Difference (dV/dt):**
The charge on the capacitor at any instant is q = CV. The charging current is I = dq/dt.
I = d(CV)/dt = C(dV/dt) (since C is constant)
Therefore, dV/dt = I/C
Given, I = 0.15 A

dV/dt = 0.15 / (80.1 × 10⁻¹²) 
dV/dt ≈ 1.87 × 10⁹ V/s

**(b) Displacement Current (I_d)**

According to Maxwell's theory, the displacement current is equal to the conduction current when a capacitor is being charged by a DC source or a constant current source.
The conduction current (I_c) is the current flowing in the wires connected to the capacitor.
Given, I_c = 0.15 A.
Therefore, the displacement current, I_d = I_c = 0.15 A.
This current exists in the space between the capacitor plates and ensures the continuity of current in the circuit.

**(c) Validity of Kirchhoff’s Junction Rule**

Kirchhoff's first rule (junction rule) states that the algebraic sum of currents entering a junction is zero. If we consider only the conduction current, the rule is not valid at the capacitor plate. For example, at the positive plate, conduction current flows *in* but no conduction current flows *out*.

However, if we include the displacement current, the rule becomes valid. At the positive plate, the conduction current (I_c) flows towards the plate, and the displacement current (I_d) flows away from the plate (into the gap). Since I_c = I_d, the total current entering the junction (the plate) is I_c, and the total current leaving is I_d. Thus, I_c - I_d = 0.

**Conclusion:** Kirchhoff’s junction rule is valid at each plate of the capacitor, provided we consider the sum of both conduction and displacement currents.

Question 2

Q2. A parallel plate capacitor (Fig. 8.7) made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V ac supply with a (angula

Accepted Answer

Here are the solutions to the three parts of the question:

Given data:
Radius of plates, R = 6.0 cm = 0.06 m
Capacitance, C = 100 pF = 100 × 10⁻¹² F = 10⁻¹⁰ F
RMS voltage, V_rms = 230 V
Angular frequency, ω = 300 rad s⁻¹

**(a) RMS value of the conduction current (I_rms)**

First, we calculate the capacitive reactance (X_C).
X_C = 1 / (ωC)
X_C = 1 / (300 rad s⁻¹ × 10⁻¹⁰ F) = 1 / (3 × 10⁻⁸) Ω ≈ 3.33 × 10⁷ Ω

Now, we can find the rms value of the conduction current using Ohm's law for AC circuits.
I_rms = V_rms / X_C
I_rms = 230 V / (3.33 × 10⁷ Ω) ≈ 6.9 × 10⁻⁶ A = 6.9 µA

**(b) Conduction current vs. Displacement current**

Yes, the conduction current is equal to the displacement current. For an ideal capacitor connected to an AC source, the conduction current flowing in the wires is equal in magnitude to the displacement current between the plates at all instants. I_c(t) = I_d(t).
Therefore, the rms value of the displacement current is also 6.9 µA.

**(c) Amplitude of magnetic field B at a point 3.0 cm from the axis**

The magnetic field between the plates is produced by the displacement current. We use the Ampere-Maxwell law for a circular loop of radius r = 3.0 cm = 0.03 m, concentric with the plates.
∮ B ⋅ dl = μ₀I_d_enclosed
Since B is constant in magnitude along the loop and parallel to dl, B(2πr) = μ₀I_d_enclosed.

The displacement current is distributed uniformly over the area of the plates. The displacement current enclosed by the loop of radius r is:
I_d_enclosed = (Total Displacement Current) × (Area of loop / Area of plate)
I_d_enclosed = I_d × (πr² / πR²) = I_d × (r/R)²

So, B(2πr) = μ₀ [I_d × (r/R)²]
B = (μ₀I_d r) / (2πR²)

To find the amplitude of B (B₀), we need the amplitude of the displacement current (I₀). The amplitude of the current is related to the rms value by I₀ = I_rms × √2.
I₀ = (6.9 × 10⁻⁶ A) × √2 ≈ 9.76 × 10⁻⁶ A

Now, substitute the values to find B₀:
B₀ = (μ₀I₀r) / (2πR²)
B₀ = (4π × 10⁻⁷ T·m/A × 9.76 × 10⁻⁶ A × 0.03 m) / (2π × (0.06 m)²)
B₀ = (2 × 10⁻⁷ × 9.76 × 10⁻⁶ × 0.03) / (0.0036)
B₀ ≈ 1.63 × 10⁻¹¹ T

The amplitude of the magnetic field at a point 3.0 cm from the axis is 1.63 × 10⁻¹¹ T.

Question 3

Q3. What physical quantity is the same for X-rays of wavelength 10⁻¹⁰ m, red light of wavelength 6800 Å, and radiowaves of wavelength 500 m?

Accepted Answer

The physical quantity that is the same for all electromagnetic waves, including X-rays, red light, and radiowaves, when they travel in a vacuum is their **speed**.

All electromagnetic waves propagate through a vacuum at the same constant speed, known as the speed of light, denoted by 'c'.

The value of this speed is approximately **c = 3 × 10⁸ m/s**.

While these waves have vastly different wavelengths (10⁻¹⁰ m, 6800 Å = 6.8 × 10⁻⁷ m, 500 m) and consequently different frequencies (since c = fλ), their speed in a vacuum remains the same.

Question 4

Q4. A plane electromagnetic wave travels in vacuum along z-direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is

Accepted Answer

This question has two parts:

**1. Directions of Electric and Magnetic Field Vectors**

For a plane electromagnetic wave, the electric field vector (E), the magnetic field vector (B), and the direction of wave propagation are mutually perpendicular to each other. This is a key characteristic of the transverse nature of electromagnetic waves.

Given that the wave travels along the z-direction:
*   The electric field vector (E) and the magnetic field vector (B) must lie in the plane perpendicular to the z-direction, which is the **x-y plane**.
*   The vectors E and B are also perpendicular to each other.
*   The direction of propagation is given by the direction of the cross product E × B. So, if E is along the x-axis, B must be along the y-axis for the wave to propagate along the z-axis (since î × ĵ = k̂).

**2. Calculation of Wavelength**

The relationship between the speed of light (c), frequency (f or ν), and wavelength (λ) is given by:
c = fλ

Given:
Frequency, f = 30 MHz = 30 × 10⁶ Hz
Speed of light in vacuum, c = 3 × 10⁸ m/s

We can find the wavelength (λ) by rearranging the formula:
λ = c / f
λ = (3 × 10⁸ m/s) / (30 × 10⁶ Hz)
λ = (3 × 10⁸) / (3 × 10⁷) m
λ = 10 m

Therefore, the wavelength of the electromagnetic wave is 10 meters.

Question 5

Q5. A radio can tune in to any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band?

Accepted Answer

To find the corresponding wavelength band, we need to calculate the wavelength (λ) for the minimum and maximum frequencies (f) in the given band. The relationship between speed (c), frequency (f), and wavelength (λ) for an electromagnetic wave is:

λ = c / f

where c is the speed of light in vacuum, c ≈ 3 × 10⁸ m/s.

**1. Wavelength for the lower frequency limit:**

Given lower frequency, f₁ = 7.5 MHz = 7.5 × 10⁶ Hz.
The corresponding wavelength is λ₁:
λ₁ = c / f₁
λ₁ = (3 × 10⁸ m/s) / (7.5 × 10⁶ Hz)
λ₁ = (300 / 7.5) m = 40 m

**2. Wavelength for the upper frequency limit:**

Given upper frequency, f₂ = 12 MHz = 12 × 10⁶ Hz.
The corresponding wavelength is λ₂:
λ₂ = c / f₂
λ₂ = (3 × 10⁸ m/s) / (12 × 10⁶ Hz)
λ₂ = (300 / 12) m = 25 m

Since wavelength is inversely proportional to frequency, the higher frequency corresponds to the shorter wavelength, and the lower frequency corresponds to the longer wavelength.

Therefore, the corresponding wavelength band is from **25 m to 40 m**.

Question 6

Q6. A plane electromagnetic wave travels in vacuum along z-direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is

Accepted Answer

### Directions of Electric and Magnetic Field Vectors
For a plane electromagnetic wave traveling in vacuum along the z-direction, the electric field vector (E) and the magnetic field vector (B) must be perpendicular to each other and also perpendicular to the direction of wave propagation (z-direction). This means that E and B lie in the x-y plane.

Specifically, if the electric field oscillates along the x-axis, then the magnetic field will oscillate along the y-axis, and vice-versa. The direction of propagation of the electromagnetic wave is given by the cross product E × B.

### Wavelength Calculation
Given:
Frequency, ν = 30 MHz = 30 × 10⁶ Hz
Speed of light in vacuum, c = 3 × 10⁸ m/s

The relationship between speed of light (c), frequency (ν), and wavelength (λ) is given by:
c = νλ

Therefore, the wavelength λ can be calculated as:
λ = c / ν
λ = (3 × 10⁸ m/s) / (30 × 10⁶ Hz)
λ = (3 × 10⁸) / (3 × 10⁷) m
λ = 10 m

The wavelength of the electromagnetic wave is 10 meters.

Question 7

Q7. A radio can tune in to any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band?

Accepted Answer

To find the corresponding wavelength band, we need to calculate the wavelength for both the lower and upper frequency limits of the radio band.

We use the fundamental relationship between the speed of light (c), frequency (ν), and wavelength (λ):
c = νλ

From this, the wavelength can be calculated as: λ = c / ν

Given:
Speed of light in vacuum, c = 3 × 10⁸ m/s

### 1. For the lower frequency limit:
ν₁ = 7.5 MHz = 7.5 × 10⁶ Hz

λ₁ = c / ν₁
λ₁ = (3 × 10⁸ m/s) / (7.5 × 10⁶ Hz)
λ₁ = (300 × 10⁶) / (7.5 × 10⁶) m
λ₁ = 300 / 7.5 m
λ₁ = 40 m

### 2. For the upper frequency limit:
ν₂ = 12 MHz = 12 × 10⁶ Hz

λ₂ = c / ν₂
λ₂ = (3 × 10⁸ m/s) / (12 × 10⁶ Hz)
λ₂ = (300 × 10⁶) / (12 × 10⁶) m
λ₂ = 300 / 12 m
λ₂ = 25 m

Therefore, the corresponding wavelength band is from 25 m to 40 m.

Question 8

Q8.10. Suppose that the electric field amplitude of an electromagnetic wave is E₀ = 120 N/C and that its frequency is ν = 50.0 MHz. (a) Determine, B₀, ω, k, and λ. (b) Find expressions fo

Accepted Answer

Given:
Electric field amplitude, E₀ = 120 N/C
Frequency, ν = 50.0 MHz = 50.0 × 10⁶ Hz
Speed of light, c = 3 × 10⁸ m/s

**(a) Determination of B₀, ω, k, and λ**

1.  **Magnetic Field Amplitude (B₀):**
    The relation is B₀ = E₀ / c.
    B₀ = (120 N/C) / (3 × 10⁸ m/s) = 40 × 10⁻⁸ T = 4.0 × 10⁻⁷ T (or 400 nT).

2.  **Angular Frequency (ω):**
    The relation is ω = 2πν.
    ω = 2π × (50.0 × 10⁶ Hz) = 100π × 10⁶ rad/s = π × 10⁸ rad/s.
    ω ≈ 3.14 × 10⁸ rad/s.

3.  **Wavelength (λ):**
    The relation is λ = c / ν.
    λ = (3 × 10⁸ m/s) / (50.0 × 10⁶ Hz) = (300 / 50) m = 6.0 m.

4.  **Wave Number / Propagation Constant (k):**
    The relation is k = 2π / λ.
    k = 2π / 6.0 m = π/3 rad/m.
    k ≈ 1.05 rad/m.
    (Alternatively, k = ω/c = (π × 10⁸ rad/s) / (3 × 10⁸ m/s) = π/3 rad/m).

**(b) Expressions for E and B**

Let's assume the wave propagates along the positive x-axis. The electric field E can be taken along the y-axis, and the magnetic field B along the z-axis, as they must be mutually perpendicular.

The general expressions for a plane electromagnetic wave are:
E(x, t) = E₀ sin(kx - ωt) ŷ
B(x, t) = B₀ sin(kx - ωt) ẑ

Substituting the calculated values:

**Expression for Electric Field (E):**
E(x, t) = 120 sin( (π/3)x - (π × 10⁸)t ) ŷ  (N/C)
Or using approximate values:
E(x, t) = 120 sin(1.05x - 3.14 × 10⁸t) ŷ  (N/C)

**Expression for Magnetic Field (B):**
B(x, t) = (4.0 × 10⁻⁷) sin( (π/3)x - (π × 10⁸)t ) ẑ  (T)
Or using approximate values:
B(x, t) = (4.0 × 10⁻⁷) sin(1.05x - 3.14 × 10⁸t) ẑ  (T)

Question 9

Q8.6. A plane electromagnetic wave travels in vacuum along z-direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is

Accepted Answer

For a plane electromagnetic wave propagating in a specific direction, the electric field (E) and magnetic field (B) vectors are mutually perpendicular to each other and also perpendicular to the direction of wave propagation.

1.  **Directions of E and B vectors**: Since the wave travels along the z-direction, the electric and magnetic field vectors must lie in the plane perpendicular to it, which is the x-y plane. They are always perpendicular to each other. For example, if the electric field vector E is along the x-axis, the magnetic field vector B will be along the y-axis, such that the direction of propagation (z-axis) is given by the direction of E × B.

2.  **Wavelength Calculation**: The relationship between frequency (ν), wavelength (λ), and the speed of light in vacuum (c) is given by c = νλ.
    Given:
    Frequency, ν = 30 MHz = 30 × 10⁶ Hz
    Speed of light in vacuum, c = 3 × 10⁸ m/s

Wavelength, λ = c / ν
    λ = (3 × 10⁸ m/s) / (30 × 10⁶ Hz)
    λ = (3/30) × 10⁸⁻⁶ m
    λ = 0.1 × 10² m
    λ = 10 m

Therefore, the electric and magnetic fields are in the x-y plane, and the wavelength of the wave is 10 meters.

Question 10

Q8.7. A radio can tune in to any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band?

Accepted Answer

To find the corresponding wavelength band, we need to calculate the wavelengths for the minimum and maximum frequencies in the given band using the relation c = νλ, where c is the speed of light in vacuum.

Given:
Frequency band: 7.5 MHz to 12 MHz
Minimum frequency, ν₁ = 7.5 MHz = 7.5 × 10⁶ Hz
Maximum frequency, ν₂ = 12 MHz = 12 × 10⁶ Hz
Speed of light, c = 3 × 10⁸ m/s

**Calculation for the wavelength (λ₁) corresponding to ν₁:**
λ₁ = c / ν₁
λ₁ = (3 × 10⁸ m/s) / (7.5 × 10⁶ Hz)
λ₁ = (300 / 7.5) m
λ₁ = 40 m

**Calculation for the wavelength (λ₂) corresponding to ν₂:**
λ₂ = c / ν₂
λ₂ = (3 × 10⁸ m/s) / (12 × 10⁶ Hz)
λ₂ = (300 / 12) m
λ₂ = 25 m

Since wavelength is inversely proportional to frequency, the lower frequency corresponds to the higher wavelength and vice-versa. Therefore, the corresponding wavelength band is from 25 m to 40 m.

Question 11

Q8.8. A charged particle oscillates about its mean equilibrium position with a frequency of 10⁹ Hz. What is the frequency of the electromagnetic waves produced by the oscillator?

Accepted Answer

The source of electromagnetic waves is an accelerated charge. An oscillating charged particle is an example of an accelerated charge, as its velocity continuously changes.

The frequency of the electromagnetic waves produced by an oscillating charged particle is exactly equal to the frequency of its oscillation.

Given:
Frequency of oscillation of the charged particle = 10⁹ Hz

Therefore, the frequency of the electromagnetic waves produced by the oscillator is also 10⁹ Hz.
This frequency falls in the microwave region of the electromagnetic spectrum.

Question 12

Q8.9. The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B₀ = 510 nT. What is the amplitude of the electric field part of the wave?

Accepted Answer

In an electromagnetic wave propagating in a vacuum, the amplitudes of the electric field (E₀) and the magnetic field (B₀) are related by the speed of light in vacuum (c).

The relation is: E₀ = c * B₀

Given:
Amplitude of the magnetic field, B₀ = 510 nT = 510 × 10⁻⁹ T
Speed of light in vacuum, c = 3 × 10⁸ m/s

Now, we can calculate the amplitude of the electric field (E₀):
E₀ = (3 × 10⁸ m/s) × (510 × 10⁻⁹ T)
E₀ = 1530 × 10⁸⁻⁹ V/m
E₀ = 1530 × 10⁻¹ V/m
E₀ = 153 V/m (or N/C)

Thus, the amplitude of the electric field part of the wave is 153 V/m.

Question 13

Q9. The peak magnetic field in an electromagnetic wave is 2.0 × 10⁻⁷ T. What is the peak electric field strength?

Accepted Answer

In an electromagnetic wave, the peak electric field strength (E₀) and the peak magnetic field strength (B₀) are related by the speed of light in vacuum (c).

The relationship is given by:
E₀ = cB₀

Given:
Peak magnetic field strength, B₀ = 2.0 × 10⁻⁷ T
Speed of light in vacuum, c = 3 × 10⁸ m/s

Substitute the given values into the formula:
E₀ = (3 × 10⁸ m/s) × (2.0 × 10⁻⁷ T)
E₀ = (3 × 2.0) × (10⁸ × 10⁻⁷) N/C
E₀ = 6.0 × 10¹ N/C
E₀ = 60 N/C

Therefore, the peak electric field strength is 60 N/C.

Question 14

Q10. An electromagnetic wave has an electric field amplitude of 120 N C⁻¹ and a frequency of 50.0 MHz. (a) Determine, the amplitude of the magnetic field. (b) Determine, the angular fre

Accepted Answer

Given:
Electric field amplitude, E₀ = 120 N C⁻¹
Frequency, ν = 50.0 MHz = 50.0 × 10⁶ Hz
Speed of light in vacuum, c = 3 × 10⁸ m/s

### (a) Amplitude of the magnetic field (B₀)
The relationship between the peak electric field (E₀) and the peak magnetic field (B₀) in an electromagnetic wave is:
E₀ = cB₀

Therefore, B₀ = E₀ / c
B₀ = (120 N C⁻¹) / (3 × 10⁸ m/s)
B₀ = 40 × 10⁻⁸ T
B₀ = 4.0 × 10⁻⁷ T

### (b) Angular frequency of the wave (ω)
Angular frequency (ω) is related to the linear frequency (ν) by the formula:
ω = 2πν

ω = 2π (50.0 × 10⁶ Hz)
ω = 100π × 10⁶ rad/s
ω ≈ 314.159 × 10⁶ rad/s
ω ≈ 3.14 × 10⁸ rad/s

### (c) Wavelength of the wave (λ)
The relationship between speed of light (c), frequency (ν), and wavelength (λ) is:
c = νλ

Therefore, λ = c / ν
λ = (3 × 10⁸ m/s) / (50.0 × 10⁶ Hz)
λ = (300 × 10⁶) / (50.0 × 10⁶) m
λ = 6 m

### (d) Wave number (k)
Wave number (k) is related to the wavelength (λ) by the formula:
k = 2π / λ

k = 2π / 6 m⁻¹
k = π / 3 m⁻¹
k ≈ 1.047 m⁻¹

Alternatively, k can also be calculated using ω and c:
k = ω / c
k = (3.14159 × 10⁸ rad/s) / (3 × 10⁸ m/s)
k ≈ 1.047 m⁻¹

Summary of results:
(a) B₀ = 4.0 × 10⁻⁷ T
(b) ω = 3.14 × 10⁸ rad/s
(c) λ = 6 m
(d) k = 1.05 m⁻¹ (rounded to two decimal places)

Question 15

Q11. Suppose that the electric field amplitude of an electromagnetic wave is E₀ = 120 N/C and that its frequency is ν = 50.0 MHz. (a) Find, B₀, ω, k, and λ. (b) Find the expressions for

Accepted Answer

Given:
Electric field amplitude, E₀ = 120 N/C
Frequency, ν = 50.0 MHz = 50.0 × 10⁶ Hz
Speed of light in vacuum, c = 3 × 10⁸ m/s

### (a) Finding B₀, ω, k, and λ

#### 1. Amplitude of the magnetic field (B₀)
The relationship between the peak electric field (E₀) and the peak magnetic field (B₀) in an electromagnetic wave is:
E₀ = cB₀

Therefore, B₀ = E₀ / c
B₀ = (120 N/C) / (3 × 10⁸ m/s)
B₀ = 40 × 10⁻⁸ T
**B₀ = 4.0 × 10⁻⁷ T**

#### 2. Angular frequency (ω)
Angular frequency (ω) is related to the linear frequency (ν) by the formula:
ω = 2πν

ω = 2π (50.0 × 10⁶ Hz)
ω = 100π × 10⁶ rad/s
**ω ≈ 3.14 × 10⁸ rad/s**

#### 3. Wavelength (λ)
The relationship between speed of light (c), frequency (ν), and wavelength (λ) is:
c = νλ

Therefore, λ = c / ν
λ = (3 × 10⁸ m/s) / (50.0 × 10⁶ Hz)
λ = (300 × 10⁶) / (50.0 × 10⁶) m
**λ = 6 m**

#### 4. Wave number (k)
Wave number (k) is related to the wavelength (λ) by the formula:
k = 2π / λ

k = 2π / 6 m⁻¹
**k = π / 3 m⁻¹ ≈ 1.05 m⁻¹**

### (b) Finding the expressions for E and B

To write the expressions for E and B, we need to assume a direction of propagation and polarization. Let's assume the electromagnetic wave is propagating in the positive x-direction. In this case, the electric field (E) and magnetic field (B) vectors will oscillate in the y-z plane.

Let's assume the electric field oscillates along the y-direction. Then the magnetic field will oscillate along the z-direction (since E, B, and direction of propagation are mutually perpendicular, and E × B gives the direction of propagation).

The general equations for a plane progressive electromagnetic wave propagating along the +x direction are:
E(x, t) = E₀ sin(kx - ωt)
B(x, t) = B₀ sin(kx - ωt)

Substituting the calculated values:
E₀ = 120 N/C
B₀ = 4.0 × 10⁻⁷ T
ω = 100π × 10⁶ rad/s
k = π / 3 m⁻¹

Therefore, the expressions for the electric and magnetic fields are:

**Electric field:**
**E(x, t) = (120 N/C) sin[(π/3 m⁻¹)x - (100π × 10⁶ rad/s)t] ŷ**

**Magnetic field:**
**B(x, t) = (4.0 × 10⁻⁷ T) sin[(π/3 m⁻¹)x - (100π × 10⁶ rad/s)t] ẑ**

(Note: ŷ and ẑ are unit vectors along the y and z axes, respectively. The choice of direction for E and B is arbitrary as long as they are perpendicular to each other and to the direction of propagation.)

Question 16

Q12. The oscillating magnetic field in a plane electromagnetic wave is given by B_y = (8 × 10⁻⁶ T) sin [2 × 10¹¹ t + 300πx] T.
(a) Calculate the wavelength of the electromagnetic wave.

Accepted Answer

The given equation for the oscillating magnetic field is B_y = (8 × 10⁻⁶ T) sin [2 × 10¹¹ t + 300πx] T.
Comparing this with the standard wave equation B_y = B₀ sin(ωt + kx):
We have:
Magnetic field amplitude, B₀ = 8 × 10⁻⁶ T
Angular frequency, ω = 2 × 10¹¹ rad/s
Wave number, k = 300π rad/m

(a) **Wavelength of the electromagnetic wave (λ):**
The wave number k is related to the wavelength λ by the formula:
k = 2π/λ
Therefore, λ = 2π/k
λ = 2π / (300π)
λ = 1/150 m
λ = 0.00667 m or 6.67 mm

(b) **Expression for the oscillating electric field:**
In an electromagnetic wave, the electric field (E) and magnetic field (B) are perpendicular to each other and also perpendicular to the direction of wave propagation. The direction of propagation is given by the sign in (ωt + kx). Since it's +kx, the wave is propagating in the negative x-direction.
If the magnetic field B is along the y-axis and the wave propagates along the negative x-axis, then the electric field E must be along the z-axis (such that E × B is in the direction of propagation, i.e., (+z) × (+y) = -x).

The amplitude of the electric field E₀ is related to the amplitude of the magnetic field B₀ by the speed of light c:
E₀ = cB₀
Where c = 3 × 10⁸ m/s (speed of light in vacuum)
E₀ = (3 × 10⁸ m/s) × (8 × 10⁻⁶ T)
E₀ = 2400 V/m

The electric field oscillates with the same frequency and phase as the magnetic field. So, the expression for the oscillating electric field will be:
E_z = E₀ sin(ωt + kx)
E_z = (2400 V/m) sin [2 × 10¹¹ t + 300πx] V/m

Question 17

Q13. How are electromagnetic waves produced? What is the source of energy of these waves?

Accepted Answer

### Production of Electromagnetic Waves
Electromagnetic waves are produced by **accelerating electric charges**.

Here's a detailed explanation:
1.  **Oscillating Charge:** When an electric charge oscillates (i.e., undergoes accelerated motion), it produces an oscillating electric field in its vicinity.
2.  **Varying Electric Field:** According to Maxwell's equations, a time-varying electric field produces a time-varying magnetic field (displacement current).
3.  **Varying Magnetic Field:** This time-varying magnetic field, in turn, produces a time-varying electric field.
4.  **Self-Sustaining Propagation:** This continuous interplay of mutually perpendicular, time-varying electric and magnetic fields, each regenerating the other, propagates through space as an electromagnetic wave. The electric and magnetic fields oscillate perpendicular to each other and also perpendicular to the direction of wave propagation.

Common examples of accelerating charges include electrons oscillating in an antenna (producing radio waves), or electrons accelerating/decelerating in X-ray tubes.

### Source of Energy of Electromagnetic Waves
The source of energy of electromagnetic waves is the **energy supplied to the accelerating charge itself**.

For instance:
*   In a radio transmitter, electrical energy is supplied to an oscillating LC circuit, which causes electrons in the antenna to oscillate. This electrical energy is converted into the energy of the electromagnetic waves radiated by the antenna.
*   When an electron is accelerated or decelerated (e.g., in an X-ray tube or during atomic transitions), the kinetic energy lost by the electron or the energy released during the transition is converted into the energy of the emitted electromagnetic radiation (X-rays, visible light, etc.).

In essence, the energy carried by the electromagnetic wave originates from the work done on the charge to make it accelerate or oscillate.

Question 18

Q14. Write the following functions of the parts of the electromagnetic spectrum:
(a) Radio waves:
(b) Microwaves:
(c) Infrared waves:
(d) Ultraviolet waves:
(e) X-rays:

Accepted Answer

Here are the functions/applications of the specified parts of the electromagnetic spectrum:

(a) **Radio waves:**
*   **Radio and Television Communication:** Used for broadcasting radio and television signals over long distances.
*   **Wireless Communication:** Essential for mobile phones, Wi-Fi, and other wireless data transmission.
*   **Remote Control:** Used in garage door openers and car key fobs.

(b) **Microwaves:**
*   **Microwave Ovens:** Used for cooking food by causing water molecules to vibrate and heat up.
*   **Radar Systems:** Used in air traffic control, weather forecasting, and speed detection (e.g., speed guns).
*   **Satellite Communication:** Used for transmitting signals to and from satellites.

(c) **Infrared waves:**
*   **Remote Controls:** Commonly used in TV remote controls.
*   **Thermal Imaging/Night Vision:** Used in night vision goggles and thermal cameras to detect heat signatures.
*   **Heating Appliances:** Used in infrared heaters, physical therapy for muscle relaxation, and solar water heaters.
*   **Optical Fibres:** Used in optical communication systems.

(d) **Ultraviolet waves:**
*   **Sterilization:** Used to kill germs and bacteria in water purifiers (UV purifiers) and medical equipment.
*   **Tanning Beds:** Used to stimulate melanin production in the skin.
*   **Forensic Applications:** Used to detect forged documents, fingerprints, and body fluids.
*   **Vitamin D Production:** Essential for the human body to produce Vitamin D.

(e) **X-rays:**
*   **Medical Imaging:** Used to visualize bones and internal structures in medical diagnostics (radiography).
*   **Security Scanners:** Used in airport security to scan luggage for hidden objects.
*   **Industrial Inspection:** Used to detect flaws or cracks in metal castings and welds.
*   **Crystallography:** Used in X-ray diffraction to study the atomic and molecular structure of crystals.

Question 19

Q15. In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 × 10¹⁰ Hz and amplitude 48 V m⁻¹.
(a) What is the wavelength of the wave?
(b) Wh

Accepted Answer

Given:
Frequency, f = 2.0 × 10¹⁰ Hz
Amplitude of electric field, E₀ = 48 V m⁻¹
Speed of light in vacuum, c = 3 × 10⁸ m/s
Permittivity of free space, ε₀ = 8.854 × 10⁻¹² C² N⁻¹ m⁻²
Permeability of free space, μ₀ = 4π × 10⁻⁷ T m A⁻¹

(a) **Wavelength of the wave (λ):**
The relationship between speed of light (c), frequency (f), and wavelength (λ) is given by:
c = fλ
Therefore, λ = c/f
λ = (3 × 10⁸ m/s) / (2.0 × 10¹⁰ Hz)
λ = 1.5 × 10⁻² m
λ = 0.015 m or 1.5 cm

(b) **Amplitude of the oscillating magnetic field (B₀):**
In an electromagnetic wave, the amplitudes of the electric and magnetic fields are related by the speed of light:
E₀ = cB₀
Therefore, B₀ = E₀/c
B₀ = (48 V m⁻¹) / (3 × 10⁸ m/s)
B₀ = 16 × 10⁻⁸ T
B₀ = 1.6 × 10⁻⁷ T

(c) **Show that the average energy density of the electric field equals the average energy density of the magnetic field (u_E_avg = u_B_avg):**

**Average energy density of the electric field (u_E_avg):**
u_E_avg = (1/2)ε₀E²_rms
Since E_rms = E₀/√2, then E²_rms = E₀²/2.
u_E_avg = (1/2)ε₀(E₀²/2)
u_E_avg = (1/4)ε₀E₀²

**Average energy density of the magnetic field (u_B_avg):**
u_B_avg = (1/2)(B²_rms/μ₀)
Since B_rms = B₀/√2, then B²_rms = B₀²/2.
u_B_avg = (1/2)(B₀² / (2μ₀))
u_B_avg = (1/4)(B₀²/μ₀)

Now, we need to show that u_E_avg = u_B_avg.
We know that E₀ = cB₀, so B₀ = E₀/c.
Substitute B₀ into the expression for u_B_avg:
u_B_avg = (1/4) * ((E₀/c)² / μ₀)
u_B_avg = (1/4) * (E₀² / (c²μ₀))

We also know the relationship between c, μ₀, and ε₀: c = 1/√(μ₀ε₀).
Therefore, c² = 1/(μ₀ε₀), which implies μ₀c² = 1/ε₀ or 1/(c²μ₀) = ε₀.

Substitute 1/(c²μ₀) = ε₀ into the expression for u_B_avg:
u_B_avg = (1/4) * E₀² * ε₀
u_B_avg = (1/4)ε₀E₀²

Comparing this with the expression for u_E_avg:
u_E_avg = (1/4)ε₀E₀²

Thus, we have shown that **u_E_avg = u_B_avg**.

Chapter 8: Electromagnetic Waves